CODEWITHSUNDEEP
javaif-statement
Mistu4u
Mistu4u

Reputation: 5416

How to check whether an int is within a range of values

In a particular if-else, I have to identify if a given value falls within a range. The condition part looks like this:

else if (a==2 && b.equalsIgnoreCase("line")
     && <<Here I need to search if c falls within a range>> && d)

where a is int, b is string, c is int and d is a boolean. Now if c falls within 1 to 8, the condition is true, else it is false. How can I do that?

Upvotes: 2

Views: 1468

Answers (5)

Sbspider
Sbspider

Reputation: 399

Unfortunately, the only way to do it would be using if else statements (as far as I know). A switch statement does not accept conditional operators, and so would not work for that. As suggested by some others, you could use seperate methods, as the methods, would allow you to check if other values were in range as well. Instead of just for this value, however, you could go for something like 

public Boolean inRange(int num, int lowBound, int topBound)
{
    return (num > lowBound && num < topBound);

}

and just use it in your code like this

else if (a==2 && b.equalsIgnoreCase("line")
     && inBound(c, 1, 8) && d)

It does not look too messy, and will work (you'll need to decide whether it is inclusive or exclusive bounds though).

The reason for the changed method is that it could be used in other places as well, making it useful for range checking.

Upvotes: 0

Alexey Grigorev
Alexey Grigorev

Reputation: 2425

Surprise surprise, it is c >= low && c <= high

To answer to the update, you'll need to employ a set

Set<Integer> validValues = new HashSet<Integer>();
validValues.add(1);
validValues.add(4);
validValues.add(9);
validValues.add(10);
validValues.add(19);

if (validValues.contains(currentVal)) {
    // do stuff
}

To curb java's verbosity you may use Guava's immutable set:

Set<Integer> validValues = ImmutableSet.of(1, 4, 9, 10, 19);

Upvotes: 3

SpiderShlong
SpiderShlong

Reputation: 224

You simply need to add:

c > 1 && c < 8

If you ever feel that your conditionals are getting too complicated I'd suggest creating Boolean methods that simplify them down a little. This is a very moderate case, but for example this range problem could be represented by:

public boolean inRange(int num)
{
    return (num > 1 && num < 8);
}

Upvotes: 0

Rahul
Rahul

Reputation: 45060

I think you need you this condition for c

(c > 1 && c < 8) // 1 and 8 are exclusive
(c => 1 && c <= 8) // 1 and 8 are inclusive

Full sample

else if (a==2 && b.equalsIgnoreCase("line")
     && (c > 1 && c < 8) && d)

If you need to check if the values belongs to a set of values, you need to use a Set and then check if the c belongs to that or not. In your original question, it was mentioned as a range and thus the answer. Anyways, this is how you can check a value in a set.

Integer[] arr = {1,4,9,11,13};
Set<Integer> set = new HashSet<>(Arrays.asList(arr));
...
else if (a==2 && b.equalsIgnoreCase("line")
     && (set.contains(c)) && d)

Upvotes: 3

sam
sam

Reputation: 120

Try like this, range1 and range2 are ranges from which you need to check b.

(c < range1 && c >  range2) ? d=true : d=false;

Upvotes: 0

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